2009 AMC 12A Problems/Problem 20
From AoPSWiki
(Redirected from 2009 AMC 10A Problems/Problem 23)
- The following problem is from both the 2009 AMC 12A #20 and 2009 AMC 10A #23, so both problems redirect to this page.
Problem
Convex quadrilateral 
has 
and 
. Diagonals 
and 
intersect at 
, 
, and 
and 
have equal areas. What is 
?
Contents |
Solution
Solution 1
Let ![[ABC]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/b/7/8b78d271c3d197e48980793383e67d5fd4a27dfc.gif)
denote the area of triangle 
. ![[AED] = [BEC]](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/1/9/b199513a588e8ad6a9e95b7d6773f794edec5fe4.gif)
, so ![[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/1/6/a/16a9455475d5be3d71d9fb2373516234991b4a36.gif)
. Since triangles 
and share a base, they also have the same height and thus 
and 
with a ratio of 
. 
.
Solution 2
Using the sine area formula on triangles 
and 
, as 
, we see that
Since 
, triangles 
and 
are similar. Their ratio is 
. Since 
, we must have 
, 
.
See also
| 2009 AMC 12A (Problems • Resources) | ||
| Preceded by Problem 19 | Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| 2009 AMC 10A (Problems • Resources) | ||
| Preceded by Problem 22 | Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
